$\left(x^{2}+1\right) \log x$
Let $I=\int\left(x^{2}+1\right) \log x d x=\int x^{2} \log x d x+\int \log x d x$
Let $I=I_{1}+I_{2} \ldots$ ...(1)
Where, $I_{1}=\int x^{2} \log x d x$ and $I_{2}=\int \log x d x$
$I_{1}=\int x^{2} \log x d x$
Taking $\log x$ as first function and $x^{2}$ as second function and integrating by parts, we obtain
$I_{1}=\log x-\int x^{2} d x-\int\left\{\left(\frac{d}{d x} \log x\right) \int x^{2} d x\right\} d x$
$=\log x \cdot \frac{x^{3}}{3}-\int \frac{1}{x} \cdot \frac{x^{3}}{3} d x$
$=\frac{x^{3}}{3} \log x-\frac{1}{3}\left(\int x^{2} d x\right)$
$=\frac{x^{3}}{3} \log x-\frac{x^{3}}{9}+\mathrm{C}_{1}$ ...(2)
$I_{2}=\int \log x d x$
Taking log x as first function and 1 as second function and integrating by parts, we obtain
$I_{2}=\log x \int 1 \cdot d x-\int\left\{\left(\frac{d}{d x} \log x\right) \int 1 \cdot d x\right\}$
$=\log x \cdot x-\int \frac{1}{x} \cdot x d x$
$=x \log x-\int 1 d x$
$=x \log x-x+\mathrm{C}_{2}$ ...(3)
Using equations (2) and (3) in (1), we obtain
$\begin{aligned} I &=\frac{x^{3}}{3} \log x-\frac{x^{3}}{9}+\mathrm{C}_{1}+x \log x-x+\mathrm{C}_{2} \\ &=\frac{x^{3}}{3} \log x-\frac{x^{3}}{9}+x \log x-x+\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right) \\ &=\left(\frac{x^{3}}{3}+x\right) \log x-\frac{x^{3}}{9}-x+\mathrm{C} \end{aligned}$